BATCH REACTOR

Example 1

The first order reaction (-rA = kCA)

A B

Calculate the time to reduce the number of moles of A to 1% of its initial value in a constant-volume batch reactor for the above reaction.

The specific reaction rate, k is 0.23 min-1.

Solution Given:

V

NkkCr

dmV

molNdmmolC

NN

k

AAA

AA

AA

)(

)()/(

01.0

min/23.0

3

3

0

Solution

min20)01.0ln(min/23.0

1

01.0ln

1ln

1

1

0

0

0

0

t

N

N

kN

N

kt

N

dN

kdt

kNdt

dN

VV

NkVr

dt

dN

A

A

A

A

N

NA

A

AA

AA

A

A

A

For batch reactor:

CONTINUOUS-STIRRED TANK REACTOR (CSTR)

Example 2

The first order reaction (-rA = kCA)

A B

Determine the CSTR volume necessary to reduce the exiting concentration to 10% of the entering concentration when the volumetric flow rate is 10 dm3/min and the specific reaction rate, k is 0.23 min-1.

Solution Given:

AA

AA

kCr

v

CC

k

min/dm 10

10.0

min/23.0

3

0

0

Solution

3

0

00

3

00000

0

dm 3.391

10.0min

23.0

10.0min

dm 10

V

C

CC

V

kC

CCv

r

vCvCV

r

FFV

A

AA

A

AA

A

AA

A

AAFor CSTR:

PLUG FLOW REACTOR (PFR)

Example 3

The first order reaction (-rA = kCA)

A B

is carried out in a tubular reactor in which the volumetric flow rate, v is constant, i.e., v = vo.

Determine the reactor volume necessary to reduce the exiting concentration to 10% of the entering concentration when the volumetric flow rate is 10 dm3/min and the specific reaction rate, k is 0.23 min-1.

Solution Given:

AA

AA

kCr

vv

CC

k

min/dm 10

10.0

min/23.0

3

0

0

Solution

A

A

VC

CA

A

A

A

AAA

AAAAA

AA

C

C

k

vV

dVC

dC

k

v

dVC

dC

k

v

kCrdV

Cdv

rdV

Cdv

dV

vCd

dV

vCd

dV

dF

rdV

dF

A

A

00

0

0

0

0

00

ln

0

For PFR:

3

0

0

3

00

dm 100

10.0ln

min

23.0

min

dm 10

ln

V

C

CV

C

C

k

vV

A

A

A

A

continue:

A reactor volume of 100 dm3 is necessary to convert 90% of species A entering into product B

for the parameters given.

EXERCISES

P1-15B

The reaction A B

is to be carried out isothermally in a continuous-flow reactor.

Calculate both the CSTR and PFR reactor volumes necessary to consume 99% of A (i.e., CA = 0.01CA0) when the entering molar flow rate is 5 mol/h, assuming the reaction rate –rA is:

a) -rA = k with k = 0.05 mol/h.dm3

b) -rA = kCA with k = 0.0001 s-1

The entering volumetric flow rate is 10 dm3/h.

Solution a) CSTR volume

.hmol/dm 05.0

mol/dm 005.001.0

mol/dm 5.0/

/hdm 10

mol/h 05.0 01.0

mol/h 5

:Given

3

3

0

3

000

3

0

0

0

kr

CC

vFC

v

FF

F

A

AA

AA

AA

A

Solution a) CSTR volume

3

3

3

3

3

3

000

0

dm 99

h.dm

mol 05.0

h

dm 01

dm

mol 5.001.0

h

dm 01

dm

mol 5.0

CSTR

AA

A

AA

V

k

vCvCV

r

FFV

3

3

00

0

dm 99

h.dm

mol 05.0

h

mol 501.0

h

mol 5

01.0

CSTR

AA

A

AA

V

V

k

FFV

r

FFV(Method 2)

(Method 1)

Solution b) CSTR volume

AAA

AA

AA

AA

A

CkCr

CC

vFC

v

FF

F

1-

3

0

3

000

3

0

0

0

s 0001.0

mol/dm 005.001.0

mol/dm 5.0/

/hdm 10

mol/h 05.0 01.0

mol/h 5

:Given

Solution a) CSTR volume

3

3

3

3

3

3

000

0

dm 750,2

dm

mol 5.001.0

h

3600 x

0001.0

h

dm 01

dm

mol 5.001.0

h

dm 01

dm

mol 5.0

CSTR

A

AA

A

AA

V

s

s

V

kC

vCvCV

r

FFV

3

3

00

0

dm 750,2

dm

mol 005.0

h

s 3600

s

0001.0

h

mol 501.0

h

mol 5

01.0

CSTR

A

AA

A

AA

V

V

kC

FFV

r

FFV

(Method 1)

(Method 2)

Solution a) PFR volume

3

3

00

0

dm 99

99.0h

mol 5

h.dm

mol 05.0

1

01.01

1

0

PFR

AA

F

FA

V

AA

V

V

FFk

V

dFk

dV

krdV

dF

A

A

(Method 1):

.hmol/dm 05.0

mol/dm 005.001.0

mol/dm 5.0/

/hdm 10

mol/h 05.0 01.0

mol/h 5

:Given

3

3

0

3

000

3

0

0

0

kr

CC

vFC

v

FF

F

A

AA

AA

AA

A

Solution a) PFR volume

3

3

3

3

000

00

0

dm 99

h.dm

mol 05.0

dm

mol 0.5 .990

h

dm10

01.0

)(1

)(

0

PFR

AA

C

CA

V

AAA

V

V

CCk

vV

vCdk

dV

krdV

vCd

dV

dF

A

A

(Method 2)

.hmol/dm 05.0

mol/dm 005.001.0

mol/dm 5.0/

/hdm 10

mol/h 05.0 01.0

mol/h 5

:Given

3

3

0

3

000

3

0

0

0

kr

CC

vFC

v

FF

F

A

AA

AA

AA

A

Solution b) PFR volume

3

0

0

3

01.00

01.000

0

dm 128

01.0ln

h

s 3600

s

0001.0

h

dm 01

ln0

0

0

000

PFR

A

A

C

CA

A

C

CA

C

CA

AF

FA

AV

AA

V

C

CV

Ck

vV

dCkC

v

r

vCd

r

dFdV

rdV

dF

A

A

A

A

A

A

A

A

AAA

AA

AA

AA

A

CkCr

CC

vFC

v

FF

F

1-

3

0

3

000

3

0

0

0

s 0001.0

mol/dm 005.001.0

mol/dm 5.0/

/hdm 10

mol/h 05.0 01.0

mol/h 5

:Given